Generating function of a geometric variable at one third

For a random variable $X$ taking non-negative integer values, the probability generating function is $G(z)=\mathbb{E}[z^X]$ for $|z|\le 1$. Derive $G$ when $X\sim\text{Geom}(p)$ (counting trials up to and including the first success, so $X\ge 1$), and evaluate it at $z=\tfrac13$ with $p=\tfrac23$.

$\tfrac14$. Factor out $pz$ and sum a geometric series in $(1-p)z$: $G(z)=pz\sum_{j\ge0}\big((1-p)z\big)^j=\dfrac{pz}{1-(1-p)z}$. With $p=\tfrac23$ and $z=\tfrac13$ this is $\dfrac{2/9}{8/9}=\tfrac14$.